Keeping breadth constant, depth of a cantilever of length $$l$$ of uniform strength loaded with uniformly distributed load w varies from zero at the free end and A. $$\frac{{2{\text{w}}}}{{\sigma {\text{b}}}} \times l{\text{ at the fixed end}}$$ B. $$\sqrt {\frac{{3{\text{w}}}}{{\sigma {\text{b}}}} \times l} {\text{ at the fixed end}}$$ C. $$\sqrt {\frac{{2{\text{w}}}}{{\sigma {\text{b}}}} \times l} {\text{ at the fixed end}}$$ D. $$\frac{{3{\text{w}}}}{{\sigma {\text{b}}}} \times l{\text{ at the fixed end}}$$

The correct answer is $\boxed{\frac{{2{\text{w}}}}{{\sigma {\text{b}}}} \times l}$ at the fixed end.

The depth of a cantilever of length $l$ of uniform strength loaded with uniformly distributed load $w$ varies from zero at the free end and $\frac{{2{\text{w}}}}{{\sigma {\text{b}}}} \times l$ at the fixed end. This is because the maximum bending moment occurs at the fixed end, and the bending moment is proportional to the depth of the beam.

The other options are incorrect because they do not take into account the variation of the depth of the beam along its length.